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Let $\alpha$ be a real number such that $1$ and $\alpha$ are rationally independent. I need a subset $U\subseteq\mathbb{R}$ such that for $x,x' \in \mathbb{R}$ and $x-x'\not \in \mathbb{Z}$, there exists $m \in \mathbb{Z}$ such that $x-m\alpha -u \in \mathbb{Z}$ for some $u \in U$ but $x' -m\alpha-u' \not \in \mathbb{Z}$ for all $u' \in U$. I feel like such a $U$ cannot exist. But I am not getting any contradiction by proceeding the assumption that such a $U$ exits. I also tried to construct one example. That also I couldnt prove at all.

budi
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    $x-x'\not\in\mathbb Z$ is not enough. You should have asked for $x-x'\ne n+m\alpha,;n,m\in\mathbb Z$. – Ivan Neretin Jun 25 '21 at 12:16
  • okay. In that case does a $U$ exists? Can you suggest en example of $U$ – budi Jun 25 '21 at 12:32
  • Sure, the set of all numbers of the form $x+n+m\alpha$ will do. – Ivan Neretin Jun 25 '21 at 15:52
  • are u varying $x$ also? Then second condition is not satisfied. I need a subset which works for any $x, x' \in \mathbb{R}$ with $x-x' \not \in \mathbb{Z}+\mathbb{Z}\alpha$? – budi Jun 25 '21 at 16:02
  • Varying? That's the opposite of what I am doing. I take a fixed $x$ and build a set $U$ which is good for it. If you want one set which is good for *any* $x,,x'$, then obviously you're out of luck. – Ivan Neretin Jun 25 '21 at 17:44

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